User talk:Dc987

Is it true that observation decreases an observer's entropy? For an idealised observer, with blank memory cells which get "filled" by observations, that might be true, but for real-world observers the reverse is probably true. E.g. observations lead to synapse development ==> more entropic. --Michael C Price 11:00, 1 October 2009 (UTC)

From the QM perspective. I think that observation increases the amount of information (bits) in the observer. The only way to get bits from the QM is to entangle something with something. Entangle electrons <01> + <10> - you have a bit sequence. Add another entanglement to say <01> and you get a longer bit sequence <010> + <011>. Think of observation as the same process as measurement. And measurement is entanglement.

Entropy is a measure of the uncertainty associated with a random variable. The random environment of the observer is the unentangled part of the universal state function. With the subjective time flow the observer can only get more entangled with the environment. So yes. The uncertainty decreases and the entropy decreases. --Dc987 19:58, 1 October 2009 (UTC)

Now the real-world observer, e.g. real-world experiences leads to synapse development. I think we need to analyze some simple case, e.g. what happens if some physicist measures the spin of an electron in some previously unknown to him state. This experience is definitely real.

From the classical perspective we have a closed system {environment, physicist}. Measuring apparatus and the electron are the part of the environment. Development of the synapse requires some low-entropy energy transfer from the environment to the physicist (and high-entropy energy back). The experience of having the measurement done results in the information transfer (ideally one bit) to the physicist. Combined result: the complexity of the physicist grows, the entropy of the physicist diminishes.

From the QM perspective, whatever physicist do or not do (he does not even have to be conscious or alive), the only thing that is changing (increasing) with the subjective time is the amount of the entanglement of the physicist with the environment. ('Disentangling' would be the equivalent of subjectively going backwards in time). Combined subjective result: the complexity of the physicist grows, the entropy of the physicist diminishes. --Dc987 22:43, 1 October 2009 (UTC)

I'm not sure. Your could argue: Entropy is uncertainty. Both before and afterwards the observer is in a definite state, so no uncertainty and no entropy (in principle, for an idealised observer). It's difficult because there are so many perspectives on entropy. You can also argue that it is only the clearing of the memory cells that increases entropy. --Michael C Price 10:12, 7 October 2009 (UTC)

Let's consider first the application of the third principle of thermodynamics that can also be re-stated using statistical mechanics with the help of the partition function, Z. Consider a thought experiment in which the photon passes through a crystal near zero degree Kelvin; for example, 1 microKelvin would be sufficient. Furthermore, assume that both detectors are also at 1 microK. One can also do exact calculations in the limit of T ---> 0 deg K. Then, assume that the light is linearly polarized, and that the polarizations for the two crystals are rerspectively 0, and 90, deg. Because the harmonic quantum oscillator assumed in the model is near, or at, zero degree Kelvin, it is only capable of the zero level vibration mode--a single energy level which is nevertheless not zero. The exponent term proposed becomes however zero or negligible, and as expected, there are no entropy effects in this case, as the entire exponential term takes on the value of 1.00. However, the difference in polarization predicts different probabilities for the photon at the two detectors that have two different polarizations (v. Feynman's Lectures on Physics for a simple quantum electrodynamics calculation for polarized photons). Thus, symmetry considerations do enter into the picture at least via polarization, even if the entropic effects are either zero or negligible. Hope this suggestion helps.Bci2 04:28, 28 January 2010 (UTC)talk

Er. Yes. If we have linearly polarized light and two detectors that have two different polarizations clearly we are going to have different probabilities. But I don't see though how this is helping in the analysis of the original setup. And in the limit of zero entropy we a priory going to have a symmetric probability distribution (if the setup is symmetric).

If the setup is asymmetric and we have some extra stuff happening on one of the paths that may affect the probabilities. After the photon passes through a half-silvered mirror we have probability amplitudes, not probabilities. And all these extra potential microstates of the 'heater' may still play a part.--Dc987 03:15, 5 March 2010 (UTC)